me":"- Deduces that if $p^2$ is even, then $p$ must also be even. B1\n- Let $p = 2m$ for some integer $m$, and substitutes this into the equation to give $(2m)^2 = 4q + 2$. M1\n- Simplifies the result to $4m^2 = 4q + 2$ and subsequently divides by 2 to get $2m^2 = 2q + 1$ (or $2(m^2 - q) = 1$). A1\n- Explains that the left-hand side $2m^2$ is even while the right-hand side $2q + 1$ is odd, which is a contradiction. A1\n- Concludes that the initial assumption is false, therefore no such integers $p$ and $q$ exist. B1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"parity_contradiction","label":"Parity Analysis (Even vs Odd)","steps":[{"type":"text","order":0,"title":"Deduce parity of p","content":"Since $p^{2}$ is even, $p$ must also be an even integer. \n\nThis is because the square of an odd number is always odd, so only an even integer can have an even square. Let's write $p$ as:\n\n$$p = 2m$$\n\nfor some integer $m$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$p^{2} = 4q + 2$"},{"latex":"$$\\htmlData{anchor=sub-lhs}{\\color{@sky}{(2m)^{2}}} = 4q + 2$"},{"latex":"$$\\htmlData{anchor=expanded-lhs}{\\color{@sky}{4m^{2}}} = 4q + 2$"}],"highlights":[{"color":"sky","style":"box","anchor":"sub-lhs"},{"color":"sky","style":"box","anchor":"expanded-lhs"}],"connections":[{"to":"expanded-lhs","from":"sub-lhs","color":"sky","label":"expand","style":"solid"}]},"order":1,"title":"Substitute and simplify","description":"We substitute $p = 2m$ into the equation $p^{2} = 4q + 2$ and expand the left-hand side."},{"type":"text","order":2,"title":"Divide by two","content":"Next, we divide both sides of the equation by $2$ to simplify it further:\n\n$$\\frac{4m^{2}}{2} = \\frac{4q + 2}{2}$$\n\nThis simplifies to:\n\n$$2m^{2} = 2q + 1$$"},{"type":"text","order":3,"title":"Analyze the parity","content":"Let's analyze the parity (evenness or oddness) of both sides of our simplified equation:\n\n* **Left-Hand Side (LHS):** $2m^{2}$ is a multiple of $2$, so it is always an **even** integer.\n* **Right-Hand Side (RHS):** $2q + 1$ is one more than a multiple of $2$, so it is always an **odd** integer.\n\nAn even integer can never equal an odd integer:"},{"type":"text","order":4,"title":"State the contradiction","content":"$$$2m^{2} \\neq 2q + 1$$\n\nThis contradiction shows that our initial assumption (that there exist integers $p$ and $q$ satisfying the equation) must be false.\n\nTherefore, the equation $p^{2} - 4q = 2$ has no integer solutions for $p$ and $q$."}]},{"id":"divisibility_by_4","label":"Divisibility by 4","steps":[{"type":"text","order":0,"title":"Establish that $p$ is even","content":"We start from the line provided in the prompt:\n\n$$p^{2} = 4q + 2$$\n\nSince $4q + 2 = 2(2q + 1)$, the right-hand side is clearly an even number. This means $p^{2}$ must also be even.\n\nBy the properties of even and odd numbers, if the square of an integer is even, the integer itself must be even. Therefore, $p$ is an even integer. \n\nWe can express $p$ as:\n\n$$p = 2m$$\n\nwhere $m$ is an integer.","highlights":[{"text":"p^2 = 4q + 2","location":"specification"},{"text":"so \$ p^2 \$ is even","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } p = \\htmlData{anchor=p-expr}{\\color{@sky}{2m}} \\text{ into } p^{2} = 4q + 2$"},{"latex":"$$(\\htmlData{anchor=p-sub}{\\color{@sky}{2m}})^{2} = 4q + 2$"},{"latex":"$$\\htmlData{anchor=lhs-expanded}{\\color{@purple}{4m^{2}}} = \\htmlData{anchor=rhs-expr}{\\color{@amber}{4q + 2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"p-expr"},{"color":"sky","style":"text-color","anchor":"p-sub"},{"color":"purple","style":"text-color","anchor":"lhs-expanded"},{"color":"amber","style":"text-color","anchor":"rhs-expr"}],"connections":[{"to":"p-sub","from":"p-expr","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute and expand","description":"Substitute $p = 2m$ into the equation and expand t

Algebraic Methods

Question 103

Algebraic Methods Questions

Algebraic Methods

Question 103

Algebraic Methods Questions